Acdream 1234 Two Cylinders 自适应辛普森

题目链接：点击打开链接

给定r1,r2

表示2个圆柱体的半径

这两个圆柱体高是正无穷，互相垂直，问相交的最大面积

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <cmath>
#define M 410
#define inf 0x3f3f3f3f
const double eps = 1e-8;
template <class T>
inline bool rd(T &ret) {
    char c; int sgn;
    if(c=getchar(),c==EOF) return 0;
    while(c!='-'&&(c<'0'||c>'9')) c=getchar();
    sgn=(c=='-')?-1:1;
    ret=(c=='-')?0:(c-'0');
    while(c=getchar(),c>='0'&&c<='9') ret=ret*10+(c-'0');
    ret*=sgn;
    return 1;
}
template <class T>
inline void pt(T x) {
    if (x <0) {
        putchar('-');
        x = -x;
    }
    if(x>9) pt(x/10);
    putchar(x%10+'0');
}
using namespace std;
double a, R, r;
double F(double x){
    return sqrt(R*R-x*x) * sqrt(r*r-x*x);
}
double simpson(double a, double b){
    double c = a + (b-a)/2.0;
    return (F(a) +4*F(c) + F(b)) * (b-a) / 6.0;
}
double asr(double a, double b, double eps, double A){
    double c = a + (b-a) / 2.0;
    double L = simpson(a, c), R = simpson(c, b);
    if(fabs(L+R-A) <= 15*eps) return L+R+(L+R-A)/15.0;
    return asr(a, c, eps/2.0, L) + asr(c, b, eps/2.0, R);
}
double asr(double a, double b, double eps){
    return asr(a, b, eps, simpson(a,b));
}
double solve(){
    return asr(0, r, eps);
}
int main(){
    while(cin>>R>>r){
        if(r>R)swap(r,R);
        printf("%.4f\n", solve() * 8.0);
    }
    return 0;
}
/*
2 3 2
0 3 2 1 1
3 0 3 2 0
2 3 0 1 0
1 2 1 0 2
1 0 0 2 0

*/